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Sagot :
Let's call the number of nickels n and the number of dimes d. Then we can create an equation summing up the two variables to 28, because there are a total of 28 coins:
n+d=28
Since a nickel is worth .05 and a dime is worth .10, if you multiply .05 by n, you'll get the value of nickels and same thing for dimes. The total value is 1.85, so we can create the equation:
.05n+.1d=1.95
Now you have a system of equations to solve, which can be solved using elimination:
n+d=28
.05n+.1d=1.95
Multiply the 2nd equation by 10 so you can get the coefficient in front of d to be 1. That way when you subtract the 2nd equation from the 1st equation, the d cancels out and you are left with the variable n, which you are trying to find
10(.05n+.1d)=1.95(10)
.5n+d=19.50
Now subtract the 2nd equation from the 1st one
n+d=28
-(.5n+d=19.50)
———————
.5n+0d=18.50
.5n=18.50
n=18.50/.5
18.5/.5=185/5=37
N=37
So you have 37 nickels
n+d=28
Since a nickel is worth .05 and a dime is worth .10, if you multiply .05 by n, you'll get the value of nickels and same thing for dimes. The total value is 1.85, so we can create the equation:
.05n+.1d=1.95
Now you have a system of equations to solve, which can be solved using elimination:
n+d=28
.05n+.1d=1.95
Multiply the 2nd equation by 10 so you can get the coefficient in front of d to be 1. That way when you subtract the 2nd equation from the 1st equation, the d cancels out and you are left with the variable n, which you are trying to find
10(.05n+.1d)=1.95(10)
.5n+d=19.50
Now subtract the 2nd equation from the 1st one
n+d=28
-(.5n+d=19.50)
———————
.5n+0d=18.50
.5n=18.50
n=18.50/.5
18.5/.5=185/5=37
N=37
So you have 37 nickels
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